A parallel-plate capacitor is formed from two 1.6 cm - diameter electrodes spaced 2.8 mm apart. The electric field strength inside the capacitor is 6.0 * 10^6 N/C. What is the charge (in nC) on each electrode?

Express your answer using two significant figures.

12 nC

Explanation:

Capacity of the parallel plate capacitor

C = ε₀ A/d

ε₀ is constant having value of 8.85 x 10⁻¹², A area of plate, d is distance between plate

Area of plate = π r²

= 3.14 x (0.8x 10⁻²) ²

= 2 x 10⁻⁴

C = (8.85 X 10⁻¹² x 2 x 10⁻⁴) / 2.8 x 10⁻³

= 7.08 x 10⁻¹³

Potential difference between plate = field strength x distance between plate

= 6 x 10⁶ x 2.8 x 10⁻³

= 16.8 x 10³ V

Charge on plate = CV

=7.08 x 10⁻¹³ X 16.8 X 10³

11.9 X 10⁻⁹ C

12 nC.